SOLUTION: I couldn't work letter C and #2, please assist.
Please check the rest of the work and let me know if it is correct.
Solutions of Quadratic Equations and their Applications
Question 193381: I couldn't work letter C and #2, please assist.
Please check the rest of the work and let me know if it is correct.
Solutions of Quadratic Equations and their Applications
1. Determine whether the following equations have a real or complex solution. Justify your answer.
a) x2 + 3x - 15 = 0
a = 1
b = 3
c = -15
Discriminant: b2-4ac = 3 2-4*1*(-15) = 69
Discriminant (69) is greater than zero. The equation has two solutions
b) x2 + x + 4 = 0
a = 1
b = 1
c = 4
Discriminant: b2-4ac = 1 2-4*1*4 = -15
Discriminant (-15) is less than zero. No solutions are defined.
c)
d) x2 – 8x + 16 = 0
a = 1
b = -8
c = 16
Discriminant: b2-4ac = 8 2-4*1*16 = 0
Discriminant (0) is zero. There is only one solution.
e) 2x2 - 3x + 7 = 0
b = -3
c = 7
Discriminant: b2-4ac = 3 2-4*2*7 = -47
Discriminant (-47) is less than zero. No solutions are defined.
f) x2 – 4x - 77 = 0
a = 1
b = -4
c = -77
Discriminant: b2-4ac = 4 2+4*1*(-77) = 324
Discriminant (324) is greater than zero. The equation has two solutions
g) 3x2 - 7x + 6 = 0
a = 3
b = -7
c = 6
Discriminant: b2-4ac = 7 2-4*3*6 = -23
Discriminant (-23) is less than zero. No solutions are defined.
h) 4x2 + 16x + 16 = 0
a = 4x
b = 16
c = 16
Discriminant: b2-4ac = 16 2-4*4*16 = 0
Discriminant (0) is zero. There is only one solution.
2. Find an equation for which -3 and 4 are solutions.
3. What type of solution do you get for quadratic equations where D < 0? Give reasons for your answer. Also provide an example of such a quadratic equation and find the solution of the equation.
You get complex solutions for b^2-4ac<0.
x^2 +4x +5 = 0
x = [-4 +/-(16-20)^1/2]/2 = -2 +/-i
4. Create a real-life situation that fits the equation (x + 3)(x - 5) = 0 and express the situation as the same equation.
The length of a rectangular field is 2 feet more than the width and the area of the field is 15 sq feet.
Here is how we arrived at the equation from real life situation -
Let the length be x, then width = x-2,
and area = length * width
so. x(x-2) = 15
so, x^2 - 2x -15 =0
Factoring we get
(x+3)(x - 5) = 0
so, x=5 (as lenth has to be positive)
hence width = 3 feet Found 2 solutions by yuendatlo, solver91311:Answer by yuendatlo(23) (Show Source):
1b. You calculated the discriminant correctly, but interpreted the results incorrectly. You said that there are no defined solutions. In fact, there are no real number solutions. However, there is a conjugate pair of complex solutions -- see your answer to #3. Even if your instructor wanted you to answer "No solution" to these, you should still say "No real number solution"
1c. Can't help you with this one. You didn't include the problem.
1d. Almost correct. You should say that there is one real number solution with a multiplicity of 2.
1e. Same answer as 1b.
1f. Correct.
1g. Same answer as 1b.
1h. Same answer as 1d.
2. If -3 and 4 are solutions, then and are the factors of the quadratic trinomial, so:
(
